Solve 12x^2-6x-18

$$12 x ^ { 2 } - 6 x - 18$$

Factor

$$6\left(2x-3\right)\left(x+1\right)$$

Solution Steps

Factor out 6.

$$6\left(2x^{2}-x-3\right)$$

Consider 2x^{2}-x-3. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-3\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite 2x^{2}-x-3 as \left(2x^{2}-3x\right)+\left(2x-3\right).

$$\left(2x^{2}-3x\right)+\left(2x-3\right)$$

Factor out x in 2x^{2}-3x.

$$x\left(2x-3\right)+2x-3$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$6\left(2x-3\right)\left(x+1\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$12x^{2}-6x-18=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 12\left(-18\right)}}{2\times 12}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 12\left(-18\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-6\right)±\sqrt{36-48\left(-18\right)}}{2\times 12}$$

Multiply -48 times -18.

$$x=\frac{-\left(-6\right)±\sqrt{36+864}}{2\times 12}$$

Add 36 to 864.

$$x=\frac{-\left(-6\right)±\sqrt{900}}{2\times 12}$$

Take the square root of 900.

$$x=\frac{-\left(-6\right)±30}{2\times 12}$$

The opposite of -6 is 6.

$$x=\frac{6±30}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{6±30}{24}$$

Now solve the equation x=\frac{6±30}{24} when ± is plus. Add 6 to 30.

$$x=\frac{36}{24}$$

Reduce the fraction \frac{36}{24} to lowest terms by extracting and canceling out 12.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{6±30}{24} when ± is minus. Subtract 30 from 6.

$$x=-\frac{24}{24}$$

Divide -24 by 24.

$$x=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and -1 for x_{2}.

$$12x^{2}-6x-18=12\left(x-\frac{3}{2}\right)\left(x-\left(-1\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$12x^{2}-6x-18=12\left(x-\frac{3}{2}\right)\left(x+1\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-6x-18=12\times \frac{2x-3}{2}\left(x+1\right)$$

Cancel out 2, the greatest common factor in 12 and 2.

$$12x^{2}-6x-18=6\left(2x-3\right)\left(x+1\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 12

$$x ^ 2 -\frac{1}{2}x -\frac{3}{2} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{1}{2} rs = -\frac{3}{2}$$

Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{4}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{1}{4} - u s = \frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{2}

$$(\frac{1}{4} - u) (\frac{1}{4} + u) = -\frac{3}{2}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{1}{16} - u^2 = -\frac{3}{2}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{3}{2}-\frac{1}{16} = -\frac{25}{16}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{4}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{1}{4} - \frac{5}{4} = -1 s = \frac{1}{4} + \frac{5}{4} = 1.500$$

Evaluate

$$6\left(2x-3\right)\left(x+1\right)$$